Question: Compute the integer $k > 2$ for which
\[\log_{10} (k - 2)! + \log_{10} (k - 1)! + 2 = 2 \log_{10} k!.\]
We can write the given equation as
\[\log_{10} (k - 2)! + \log_{10} (k - 1)! + \log_{10} 100 = \log_{10} (k!)^2.\]Then
\[\log_{10} [100 (k - 2)! (k - 1)!] = \log_{10} (k!)^2,\]so $100 (k - 2)! (k - 1)! = (k!)^2.$  Then
\[100 = \frac{k! \cdot k!}{(k - 2)! (k - 1)!} = k(k - 1) \cdot k = k^3 - k^2.\]So, $k^3 - k^2 - 100 = 0,$ which factors as $(k - 5)(k^4 + 4k + 20) = 0.$  The quadratic factor has no integer roots, so $k = \boxed{5}.$